The energy required to ionize potassium is 419 kj/mol
What minimum frequency of light is required to ionize potassium?
please help, thanks
We find the work function per atom e = E/N = 419000/6.023E23 = hf = photon energy; solve for the frequency, using N = 6.023E23 atoms per mole and Planck’s h = 6.63E-34 J.s:
f = E/Nh = 419000/(6.023E23*6.63E-34) = 1.04927E+15 = 1.05E15 cps. ANS.
Ionization energy = 419kJ/mol = 419000J/mol (since 1kJ = 100J)
= 419000/ 6.023*10^23 J/atom (since 1 mol = 6.023*10^23molecules) = 69.56*10^-20 J/molecule
Energy of a photon = hn (where n is the frequency). Equating the two,
6.65610^-34n = 69.56*10^-20
=> n = 1.0.45*10^15 s^-1